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* Hướng dẫn giải

Xét tích phân \({{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{f\left( \sin x \right)\cos x\text{d}x}\). Đặt \(t=sinx\Rightarrow \text{d}t=\cos x\text{d}x\)

Đổi cận

Ta có \({{I}_{1}}=\int\limits_{0}^{1}{f\left( t \right)\text{d}t=}\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( 5-x \right)\text{d}x=}\left. \left( 5x-\frac{{{x}^{2}}}{2} \right) \right|_{0}^{1}=\frac{9}{2}\)

Xét tích phân \({{I}_{2}}=\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}\). Đặt \(t=3-2x\Rightarrow \text{d}t=-2\text{d}x\Rightarrow \text{d}x=\frac{-\text{d}t}{2}\)

Đổi cận

Ta có \({{I}_{2}}=\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}=\frac{1}{2}\int\limits_{1}^{3}{f\left( t \right)\text{d}t=}\frac{1}{2}\int\limits_{1}^{3}{f\left( x \right)\text{d}x=}\frac{1}{2}\int\limits_{1}^{3}{\left( {{x}^{2}}+3 \right)\text{d}x=}\frac{1}{2}\left. \left( \frac{{{x}^{3}}}{3}+3x \right) \right|_{1}^{3}=\frac{1}{2}\left( 18-\frac{10}{3} \right)=\frac{22}{3}\)

Vậy \(I=2\int\limits_{0}^{\frac{\pi }{2}}{f\left( \sin x \right)\cos x\text{d}x+3\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}}=9+22=31\).